LAB$06$ex$1.$m

clear all; $\%$this deletes all variables

omega$0=3$;$c=2$; omega $=3\mathrm{.}6$;

param $=[$omega$0,$ c$,$ omega$]$;

$t0=0$;$yO=0$;$v0=0$;$YO=[y0$;$v0]$;$tf=40$;

options $=$ odeset$(\text{'}$AbsTol$\text{'},1$e$-10,$'relTol', $1$e$-10)$;

$,$ options, param When executing this program we get the plot in Figure $1.$ and the following output in the

command window:

computed amplitude of forced oscillation $=0\mathrm{.}1217$

theoretical amplitude $=0\mathrm{.}1217$

Figure $1$: Forced oscillation.

Lines $10-14$ deserve some explanation. Line $10$ defines a time t$1$ after which we think the

contribution of the first term in $(2)$ has become negligible compared to the second term. This

depends of course on the parameter values, in particular $c.$ With $c=2$ we obtain $e^{-\frac{1}{2}ct}$~~

$2\mathrm{.}3{10}^{-6}$ for $t=13,$ so this is certainly small enough compared to the amplitude seen on

Figure $1.$ The index $i$ of time values larger than $t1$ is then determined. The quantity $Y(i,1)$

refers to the values of $y$ associated to times larger than t$1$ only. The computed amplitude is$.$ simply half the difference between the max and the min values. This value is compared to the

theoretical value $(3).$

$($a$)$ What is the period of the forced oscillation? What is the numerical value $($modulo

$2)$ of the angle $$ defined by $(4)?$

$($b$)$ In this question you are asked to modify the file LAB$06$ex$1.$m in order to plot the

complementary solution of $(1),$ that is$,$ the first term in $(2).$ First define in the

file the angle $($alpha$)$ using $(4),$ then evaluate the complementary solution yc by

subtracting the quantity Ccos$(t-)$ from the numerical solution y$.$ Plot the

resulting quantity. Does it look like an exponentially decreasing oscillation? Why or

why not? Include the modified M$-$file and the corresponding plot.